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Bzoj 3942: [Usaco2015 Feb]Censoring(kmp)

2019-11-11 05:17:54
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3942: [Usaco2015 Feb]Censoring Description Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of material to read while waiting around in the barn during milking sessions. Unfortunately, the latest issue contains a rather inapPRopriate article on how to cook the perfect steak, which FJ would rather his cows not see (clearly, the magazine is in need of better editorial oversight). FJ has taken all of the text from the magazine to create the string S of length at most 10^6 characters. From this, he would like to remove occurrences of a substring T to censor the inappropriate content. To do this, Farmer John finds the first occurrence of T in S and deletes it. He then repeats the process again, deleting the first occurrence of T again, continuing until there are no more occurrences of T in S. Note that the deletion of one occurrence might create a new occurrence of T that didn’t exist before. Please help FJ determine the final contents of S after censoring is complete 有一個S串和一個T串,長度均小于1,000,000,設當前串為U串,然后從前往后枚舉S串一個字符一個字符往U串里添加,若U串后綴為T,則去掉這個后綴繼續流程。 Input The first line will contain S. The second line will contain T. The length of T will be at most that of S, and all characters of S and T will be lower-case alphabet characters (in the range a..z). Output The string S after all deletions are complete. It is guaranteed that S will not become empty during the deletion process. Sample Input whatthemomooofun moo Sample Output whatthefun HINT Source Silver

/*kmp+棧模擬.先將t串自匹配.然后將s與t串匹配.又是fail數組的套路題orz. */#include<cstring>#include<cstdio>#define MAXN 1000001using namespace std;int l1,l2,top,next[MAXN],fail[MAXN];char s1[MAXN],s2[MAXN],ans[MAXN];void slove(){ int p; for(int i=2;i<=l1;i++) { p=next[i-1]; while(p&&s2[i]!=s2[p+1]) p=next[p]; if(s2[i]==s2[p+1]) p++; next[i]=p; }}void kmp(){ int p; for(int i=1;i<=l1;i++) { ans[++top]=s1[i]; p=fail[top-1]; while(p&&ans[top]!=s2[p+1]) p=next[p]; if(ans[top]==s2[p+1]) p++; fail[top]=p; if(p==l2) top-=l2; } for(int i=1;i<=top;i++) printf("%c",ans[i]);}int main(){ scanf("%s",s1+1);l1=strlen(s1+1); scanf("%s",s2+1);l2=strlen(s2+1); slove();kmp(); return 0;}
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